análisis-matemático y topología-algebraica-medida y análisis-real y economía

Teorema:

lim[n = oo][ int[x = 0]-[n][ ( 1+(x/n) )^{n}·e^{(-1)·2x} ]d[x] ] = 1

Teorema:

lim[n = oo][ int[x = 0]-[n][ ( 1+(-1)·( (2x)/n ) )^{n}·e^{x} ]d[x] ] = 1

Teorema:

lim[n = oo][ int[x = 0]-[n][ ( 1+(1/p)·(x/n) )^{n}·e^{(-1)·( 1+(1/p) )·x} ]d[x] ] = 1

Teorema:

lim[n = oo][ int[x = 0]-[n][ ( 1+(-1)·(1/p)·(x/n) )^{n}·e^{( (-1)+(1/p) )·x} ]d[x] ] = 1

Teorema:

lim[n = oo][ int[x = 0]-[ln(n^{m+1})][ ( nx^{m}/(1+n) )·e^{(-x)} ]d[x] ] = m!

Teorema:

lim[n = oo][ int[x = 0]-[ln(n^{(1/m)+1})][ ( nx^{(1/m)}/(1+n) )·e^{(-x)} ]d[x] ] = m

Teorema de Hôpital-Garriga:

Teorema:

Si lim[x = 0][ g(x) ] = 0 ==>

lim[x = 0][ ( f(x) /o(x)o/ g(x) ) ] = lim[x = 0][ ( f(x) /o(x)o/ d_{x}[g(x)] )·(1/d[x]) ]

Si lim[x = oo][ g(x) ] = oo ==>

lim[x = oo][ ( f(x) /o(x)o/ g(x) ) ] = lim[x = oo][ ( f(x) /o(x)o/ d_{x}[g(x)] )·d[x] ]

Demostración:

g(x) = 0

lim[h = 0][ g(x+h)+(-1)·g(x) ] = lim[h = 0][ g(x+h)+(-0) ] = g(x)

g(x) = oo

lim[h = 0][ (1/h)^{2}·( g(x+h)+(-1)·g(x) ) ] = oo·d_{x}[g(x)] = g(x)

Teorema: [ de la integral de Euler ]

Sea n >] 1 ==>

int[x = 0]-[oo][ e^{(-1)·x^{n}} ]d[x] = (-1)·(0/n!)+(1/n!) = (1/n!)

Teorema: [ de la integral de Euler-Garriga ]

Sea n >] 1 ==>

int[x = 0]-[oo][ ( 1/(1+x^{2n}) ) ]d[x] = (pi/2)·(1/n!)+(-1)·( 0·(1/n!) ) = (pi/2)·(1/n!)

Definición: [ medida de densidad-integral ]

M( A,f(x) ) = A·int[ f(x) ]d[x]

M( (-A),f(x) ) = (-A)·int[ f(x) ]d[x]

Teorema:

M( 0,f(x) ) = int[ f(x) ]d[x]·0

Teorema:

M( (-0),f(x) ) = (-1)·int[ f(x) ]d[x]·0

Teorema:

Si A [< B ==>

M( max{A,B},f(x) ) = M( A,f(x) )+M( B,f(x) )

<==>

M( A,f(x) ) = int[ f(x) ]d[x]·0

Demostración:

( B+(-B) )·int[ f(x) ]d[x] = int[ f(x) ]d[x]·0

Teorema:

Si (-B) [< (-A) ==>

M( min{(-A),(-B)},f(x) ) = M( (-A),f(x) )+M( (-B),f(x) )

<==>

M( (-A),f(x) ) = (-1)·int[ f(x) ]d[x]·0

Demostración:

(-1)·( B+(-B) )·int[ f(x) ]d[x] = (-1)·int[ f(x) ]d[x]·0

Definición: [ medida de densidad-diferencial ]

W( A,f(x) ) = A·d_{x}[ f(x) ]

W( (-A),f(x) ) = (-A)·d_{x}[ f(x) ]

Teorema:

W( 0,f(x) ) = d_{x}[ f(x) ]·0

Teorema:

W( (-0),f(x) ) = (-1)·d_{x}[ f(x) ]·0

Teorema:

Si A [< B ==>

W( max{A,B},f(x) ) = W( A,f(x) )+W( B,f(x) )

<==>

W( A,f(x) ) = d_{x}[ f(x) ]·0

Demostración:

( B+(-B) )·d_{x}[ f(x) ] = d_{x}[ f(x) ]·0

Teorema:

Si (-B) [< (-A) ==>

W( min{(-A),(-B)},f(x) ) = W( (-A),f(x) )+W( (-B),f(x) )

<==>

W( (-A),f(x) ) = (-1)·d_{x}[ f(x) ]·0

Demostración:

(-1)·( B+(-B) )·d_{x}[ f(x) ] = (-1)·d_{x}[ f(x) ]·0

Medidas Continuas:

Teorema:

[As][ s > 0 ==> [Ed][ d > 0 & ( Si H(B) < d ==> M( H(B),f(x) ) < s ) ] ]

[A(-s)][ (-s) < 0 ==> [E(-d)][ (-d) < 0 & ( Si H(-B) > (-d) ==> M( H(-B),f(x) ) > (-s) ) ] ]

Demostración:

Sea H(B) = A ==>

M( A,f(x) ) = A·int[ f(x) ]d[x] = int[ f(x) ]d[x]·0 < s

Sea H(-B) = (-A) ==>

M( (-A),f(x) ) = (-A)·int[ f(x) ]d[x] = (-1)·int[ f(x) ]d[x]·0 > (-s)

Teorema:

[As][ s > 0 ==> [Ed][ d > 0 & ( Si H(B) < d ==> W( H(B),f(x) ) < s ) ] ]

[A(-s)][ (-s) < 0 ==> [E(-d)][ (-d) < 0 & ( Si H(-B) > (-d) ==> W( H(-B),f(x) ) > (-s) ) ] ]

Demostración:

Sea H(B) = A ==>

W( A,f(x) ) = A·d_{x}[ f(x) ] = d_{x}[ f(x) ]·0 < s

Sea H(-B) = (-A) ==>

W( (-A),f(x) ) = (-A)·d_{x}[ f(x) ] = (-1)·d_{x}[ f(x) ]·0 > (-s)

Teorema:

Sea A = x ==> d_{x}[ M( x,f(x) ) ] = M( 1,f(x) )+M( x,d_{x}[f(x)] )

Sea A = (-x) ==> d_{x}[ M( (-x),f(x) ) ] = M( (-1),f(x) )+M( (-x),d_{x}[f(x)] )

Teorema:

Sea A = x ==> d_{x}[ W( x,f(x) ) ] = W( 1,f(x) )+W( x,d_{x}[f(x)] )

Sea A = (-x) ==> d_{x}[ W( (-x),f(x) ) ] = W( (-1),f(x) )+W( (-x),d_{x}[f(x)] )

Definición:

M( A,f(x_{1},...,x_{n}) ) = A·int-...[n]...-int[ f(x_{1},...,x_{n}) ]d[x_{1}]...d[x_{n}]

M( (-A),f(x_{1},...,x_{n}) ) = (-A)·int-...[n]...-int[ f(x_{1},...,x_{n}) ]d[x_{1}]...d[x_{n}]

Definición:

W( A,f(x_{1},...,x_{n}) ) = A·d_{x_{1}...x_{n}}^{n}[ f(x_{1},...,x_{n}) ]

W( (-A),f(x_{1},...,x_{n}) ) = (-A)·d_{x_{1}...x_{n}}^{n}[ f(x_{1},...,x_{n}) ]

Examen:

Demostrad que son medidas continuas.

Teorema:

Si [Ea][Ax][ 0 [< f(x) [< a ] ==> f(x) es continua a x = a.

Demostración:

| f(x)+(-1)·f(a) | [< | a+(-a)+h | [< |x+(-a)|+|h| < 2h < s

Teorema:

Si [Ea][Ax][ 1 [< ( f(x) )^{(1/n)} [< a ] ==> f(x) es continua en x = a.

Demostración:

| f(x)+(-1)·f(a) | [< | a^{n}+(-1)·a^{n}+h | = ...

... | a^{n}+(-1)·a^{n} |+|h| = |a+(-a)|·|P(a)|+|h| [< |x+(-a)|·|P(a)|+|h| = ( |P(a)|+1 )·h < s

Teorema:

Si [Ea][Ax][ 0 [< f(x) [< |a| ] ==> f(x) es continua en x = a.

Demostración:

| f(x)+(-1)·f(a) | [< | |a|+(-1)·|a|+h | = | |a|+(-1)·|a| |+|h| = |a+(-a)|+|h| [< |x+(-a)|+|h| = 2h < s

Teorema:

Si F(x) es continua ==> f(x) es integrable Newton-Riemann.

Demostración:

| int[x = x]-[x+h][ f(x) ]d[x] | = | F(x+h)+(-1)·F(x) | < s

Teorema:

Si F(x) es continua ==> f(x) es integrable Newton-Lebesgue.

Demostración:

| lim[n = oo][ int[x = x]-[x+h][ f_{n}(x) ]d[x] ] | = | lim[n = oo][ F_{n}(x+h)+(-1)·F_{n}(x) ] | = ...

... | F(x+h)+(-1)·F(x) | < s

Teorema:

Si F(x) es monótona ==> f(x) es integrable Newton-Riemann.

Demostración:

| int[x = x]-[x+h][ f(x) ]d[x] | = | F(x+h)+(-1)·F(x) | = 0+h = h < s

Teorema:

Si F(x) es monótona ==> f(x) es integrable Newton-Lebesgue.

Demostración:

| lim[n = oo][ int[x = x]-[x+h][ f_{n}(x) ]d[x] ] | = | lim[n = oo][ F_{n}(x+h)+(-1)·F_{n}(x) ] | = ...

... | F(x+h)+(-1)·F(x) | = 0+h = h < s

Definición: [ de integral continua de Newton-Leibniz ]

Sea S( f(x),x ) = a·f(x)·x ==>

f(x)·x = int[ S( f(x+h),(x+h) )+(-1)·S( f(x),x ) ]

[As][ s >0 ==> [Ed][ d > 0 & ( Si h < d ==> | S( f(x+h),(x+h) )+(-1)·S( f(x),x ) | < s ) ] ]

Teorema:

Si f(x) es continua ==> f(x) es integrable continua Newton-Leibniz.

Demostración:

| S( f(x+h),(x+h) )+(-1)·S( f(x),x ) | = |a|·| f(x+h)·(x+h)+(-1)·f(x)·x | = ...

... |a|·| ( f(x+h)+(-1)·f(x) )·x+f(x+h)·h | [< ...

... |a|·| ( f(x+h)+(-1)·f(x) )·x |+|f(x+h)·h| = |a|·( |x|+|f(x+h)| )·h < s

Teorema:

Si f(x) es monótona ==> f(x) es integrable continua Newton-Leibniz.

Demostración:

| S( f(x+h),(x+h) )+(-1)·S( f(x),x ) | = |a|·| f(x+h)·(x+h)+(-1)·f(x)·x | = ...

... |a|·| ( f(x+h)+(-1)·f(x) )·x+f(x+h)·h| [< |a|·| ( f(x+h)+(-1)·f(x) )·x |+|f(x+h)·h| = ...

... |a|·( |(0+h)·x|+|f(x+h)·h| ) = |a|·( |x|+|f(x+h)| )·h < s

Teorema:

f(x) = x^{n} es integrable continua Newton-Leibniz.

Demostración:

S(x^{n},x) = x^{n+1}

| S( (x+h)^{n},(x+h) )+(-1)·S( x^{n},x ) | = | (x+h)^{n+1}+(-1)·x^{n+1} | = ...

... | P(x^{n+(-k)},h^{k}) |·h < s

Teorema:

f(x) = e^{nx} es integrable continua Newton-Leibniz.

Demostración:

S(e^{nx},x) = e^{nx}·x

| S( e^{n·(x+h)},(x+h) )+(-1)·S( e^{nx},x ) | = | e^{n·(x+h)}·(x+h)+(-1)·e^{nx}·x | = ...

... | e^{nx}·( e^{nh}+(-1) )·x+e^{n·(x+h)}·h | = | e^{nx}·( nh+P( (nh)^{n} ) )·x+e^{n·(x+h)}·h | = ...

... | ne^{nx}( 1+P( (nh)^{n} ) )·x+e^{n·(x+h)} |·h < s

Teorema:

f(x) = ln(x) es integrable continua Newton-Leibniz.

Demostración:

S(ln(x),x) = ln(x)·x

| S( ln(x+h),(x+h) )+(-1)·S( ln(x),x ) | = | ln(x+h)·(x+h)+(-1)·ln(x)·x | = ...

... | ln(1+(h/x))·x+ln(x+h)·h | = | (h/x)·( 1+P( (h/x)^{n} ) )·x+ln(x+h)·h | = ...

... | ( 1+P( (h/x)^{n} ) )+ln(x+h) |·h < s

Teorema:

f(x) = sin(x) es integrable continua Newton-Leibniz.

Demostración:

S(sin(x),x) = sin(x)·x

| S( sin(x+h),(x+h) )+(-1)·S( sin(x),x ) | = | sin(x+h)·(x+h)+(-1)·sin(x)·x | = ...

... | ( sin(x+h)+(-1)·sin(x) )·x+sin(x+h)·h | = ...

... | ( sin(x)·P(h^{n+(-1)})·h+cos(x)·h )·x+sin(x+h)·h | = ...

... | ( sin(x)·P(h^{n+(-1)})+cos(x) )·x+sin(x+h) |·h < s

Economía de objetos de empresa:

Lema: [ de la ventana ]

F(x,y) = 2x+2y+(-h)·( xy+(-1)·ab )

h = ( (a+b)/(ab) )

F(a,b) = 2a+2b

G(x,y) = 2x+2y+(-h)·xy

G(a,b) = h·ab

Disertación:

d_{x}[F(x,y)] = 2+(-h)·y = 0

d_{y}[F(x,y)] = 2+(-h)·x = 0

2x+(-h)·xy = 0x = 0

2y+(-h)·yx = 0y = 0

a+b = h·ab

h = ( (a+b)/(ab) )

Lema: [ de la puerta ]

F(x,y) = x+2y+(-h)·( xy+(-1)·ab )

h = ( ((a/2)+b)/(ab) )

F(a,b) = a+2b

G(x,y) = x+2y+(-h)·xy

G(a,b) = h·ab

Disertación:

d_{x}[F(x,y)] = 1+(-h)·y = 0

d_{y}[F(x,y)] = 2+(-h)·x = 0

x+(-h)·xy = 0x = 0

2y+(-h)·yx = 0y = 0

a+2b = 2h·ab

h = ( ((a/2)+b)/(ab) )

Lema: [ de la caja con tapa ]

F(x,y,z) = 2xy+2yz+2zx+(-h)·( xyz+(-1)·abc )

h = ( ((4/3)·(ab)+(4/3)·(bc)+(4/3)·(ca))/(abc) )

F(a,b,c) = 2ab+2bc+2ca

G(x,y,z) = 2xy+2yz+2zx+(-h)·xyz

G(a,b,c) = (1/2)·h·abc

Disertación:

d_{x}[F(x,y,z)] = 2y+2z+(-h)·yz = 0

d_{y}[F(x,y,z)] = 2x+2z+(-h)·zx = 0

d_{z}[F(x,y,z)] = 2y+2x+(-h)·xy = 0

2xy+2xz+(-h)·xyz = 0x = 0

2yx+2yz+(-h)·xyz = 0y = 0

2zy+2zx+(-h)·xyz = 0z = 0

4ab+4bc+4ca = 3h·abc

h = ( ((4/3)·(ab)+(4/3)·(bc)+(4/3)·(ca))/(abc) )

Lema: [ de la caja sin tapa ]

F(x,y,z) = xy+2yz+2zx+(-h)·( xyz+(-1)·abc )

h = ( ((2/3)·(ab)+(4/3)·(bc)+(4/3)·(ca))/(abc) )

F(a,b,c) = ab+2bc+2ca

G(x,y,z) = xy+2yz+2zx+(-h)·xyz

G(a,b,c) = (1/2)·h·abc

Disertación:

d_{x}[F(x,y,z)] = y+2z+(-h)·yz = 0

d_{y}[F(x,y,z)] = x+2z+(-h)·zx = 0

d_{z}[F(x,y,z)] = 2y+2x+(-h)·xy = 0

xy+2xz+(-h)·xyz = 0x = 0

yx+2yz+(-h)·xyz = 0y = 0

2zy+2zx+(-h)·xyz = 0z = 0

2ab+4bc+4ca = 3h·abc

h = ( ((2/3)·(ab)+(4/3)·(bc)+(4/3)·(ca))/(abc) )

Lema:

F(x,y) = 2x+2y+(-h)·( xy+(-1)·2! )

F(1,2) = 6

h = (3/2)

Disertación:

Laboratorio de problemas

Lema:

F(x,y,z) = 2xy+2yz+2zx+(-h)·( xyz+(-1)·3! )

F(1,2,3) = 22

h = (4/3)·(11/6) = (22/9)

Disertación:

Laboratorio de problemas.

Lema:

F(x,y) = x+2y+(-h)·( xy+(-1)·2! )

F(2,1) = 4

h = 1

Disertación:

Laboratorio de problemas.

Lema:

F(x,y,z) = xy+2yz+2zx+(-h)·( xyz+(-1)·3! )

F(3,2,1) = 16

h = (2/3)·(16/6) = (16/9)

Disertación:

Laboratorio de problemas.

Teorema:

lim[n = oo][ ( sum[k = 1]-[n][ (1/k) ]/ln(n) ) ] = 1

sum[k = 1]-[oo][ (1/k) ] = ln(2)·oo

Demostración:

lim[n = oo][ ( 1/( n·( ln(n+1)+(-1)·ln(n) )+ln(n+1)+(-1)·ln(n) ) ) ] = ( 1/(1+0) ) = 1

Teorema:

lim[n = oo][ ( sum[k = 1]-[n][ ln(k) ]/( (1/e)·n·ln(n) ) ) ] = 1

sum[k = 1]-[oo][ ln(k) ] = (1/e)·ln(2)·oo^{2} [< (1/2)·oo^{2} = sum[k = 1]-[oo][ k ]

Demostración: [ por destructor en Stolz ]

ln(e) = 1

lim[n = oo][ e·( ln(n+1)/( n·( ln(n+1)+(-1)·ln(n) )+ln(n+1) ) ) ] = e·( ln(oo)/( 1+ln(oo) ) ) = e = ln(e) = 1

Teorema:

ln(2)+(-1)·ln(2) = 0^{2}

Demostración:

ln(oo) < oo

ln(2) < 1 

ln(oo)+(-1)·ln(oo) = oo·( ln(2)+(-1)·ln(2) ) = lim[n = oo][ ln(1+(p/n)) ] = 0

Constante de Euler:

lim[n = oo][ ( sum[k = 1]-[n][ (1/k) ] )^{2}+(-1)·( ln(n) )^{2} ] = ln(2)

Teorema infinitorum de Euler:

lim[n = oo][ ( sum[k = 1]-[n][ (1/k) ] )^{2}+(-1)·( n·ln(2+(1/n)) )^{2} ] = ln(2)

Demostración:

( ln(2)·oo )^{2}+(-1)·( ln(2)·oo )^{2} = ln(2)·oo^{2}·( ln(2)+(-1)·ln(2) ) = ln(2)

( ln(2)·oo )^{2}+(-1)·( ln(2)·oo )^{2} = ln(2)·oo·( ln(2)·oo+(-1)·ln(2)·oo ) = ln(2)

( ln(2)·oo )^{2}+(-1)·( ln(2)·oo )^{2} = ln(2)·( ln(2)·oo^{2}+(-1)·ln(2)·oo^{2} ) = ln(2)

Número de Euler inverso:

lim[n = oo][ sum[k = 1]-[n][ ln(k) ]+(-1)·(1/e)·n·ln(n) ] = (1/e)

Teorema infinitorum de Euler-Garriga:

lim[n = oo][ sum[k = 1]-[n][ ln(k) ]+(-1)·(1/e)·n^{2}·ln(2+(1/n)) ] = (1/e)

Demostración:

( (1/e)·ln(2)·oo^{2} )+(-1)·( (1/e)·ln(2)·oo^{2} ) = (1/e)·oo^{2}·( ln(2)+(-1)·ln(2) ) = (1/e)

( (1/e)·ln(2)·oo^{2} )+(-1)·( (1/e)·ln(2)·oo^{2} ) = (1/e)·oo·( ln(2)·oo+(-1)·ln(2)·oo ) = (1/e)

( (1/e)·ln(2)·oo^{2} )+(-1)·( (1/e)·ln(2)·oo^{2} ) = (1/e)·( ln(2)·oo^{2}+(-1)·ln(2)·oo^{2} ) = (1/e)

Arte:

[En][ ( 1/(n+(-1)) )^{2}·( sum[k = 1]-[n][ ln(k+1) ]+(-1)·n·ln(n+1) ) = 1 ]

[En][ ( 1/(n+(-1)) )^{2}·( sum[k = 1]-[n][ ln((1/k)+1) ]+(-1)·n·ln(n+1) ) = 1 ]

Exposición:

n = 1

f(n) = 1

u(k) = 1

v(1/k) = 1

Crackeador:

Mov si,cs

Ciclo-de-teclado-positivo

Mov ax,[si]

Xor al,codigo[Interrupción-de-teclado-positiva]

Jz Condicional-de-teclado-positivo

Inc si

Jmp Ciclo-de-teclado-positivo

Condicional-de-teclado-positivo

Ciclo-Jz

Mov ax,[si]

Xor al,codigo[Jz]

Jz Condicional-Jz

Inc si

Jmp Ciclo-Jz

Condicional-Jz

Mov al,codigo [Jmp]

Mov [si],al

Mov di,not(cs)

Ciclo-de-teclado-negativo

Mov dx,[di]

Sys dl,codigo[Interrupción-de-teclado-negativa]

Jf Condicional-de-teclado-negativo

Dec di

Jmp Ciclo-de-teclado-negativo

Condicional-de-teclado-negativo

Ciclo-Jf

Mov dx,[di]

Sys dl,codigo[Jf]

Jf Condicional-Jf

Dec di

Jmp Ciclo-Jf

Condicional-Jf

Mov dl,codigo[Jmp]

Mov [di],dl

Ley:

No puede ser prójimo la demostración al teorema,

por el buey del prójimo,

y no se tiene energía.

Puede ser próximo la demostración al teorema,

por el buey del próximo,

y se tiene energía.

Teorema:

d_{x}[1] = 0

Demostración

0+0 = 0+(-0) = 0^{2}

(1/h)·( (x+h)^{0}+(-1)·x^{0} ) = (1/h)·( x^{0}+0·(h+1+0)+(-1)·x^{0} ) = ...

... (1/h)·( 0h+0^{3} ) = 0·(h/h) = 0

Teorema:

2k·0^{n} = 0^{n+1}

(2k+1)·0^{n} = 0^{n+1}+0^{n} = 0^{n}

Demostración:

0^{n}+0^{n} = 0^{n}+(-0)·0^{n+(-1)} = 0^{n+1}

(2k+2)·0^{n} = 0^{n+1}+( 0^{n}+0^{n} ) = ( 0^{n+1}+0^{n} )+0^{n} = 0^{n}+0^{n} = 0^{n+1}

Teorema:

( (2k)/m )·0^{n} = (1/m)·0^{n+1}

( (2k+1)/m )·0^{n} = (1/m)·0^{n}

Teorema:

lim[n = oo][ ( ( (2k)·n^{p}+a )/( mn^{p+k}+b ) ) ] = (1/m)·0^{k+1}

lim[n = oo][ ( ( (2k+1)·n^{p}+a )/( mn^{p+k}+b ) ) ] = (1/m)·0^{k}

Teorema:

lim[n = oo][ ( ( 13n^{p}+30 )/( 17n^{p+k}+30 ) ) ] = (1/17)·0^{k}

lim[n = oo][ ( ( 11n^{p}+30 )/( 19n^{p+k}+30 ) ) ] = (1/19)·0^{k}

Teorema:

lim[n = oo][ ( ( 8n^{p}+10 )/( 2n^{p+k}+10 ) ) ] = (1/2)·0^{k+1}

lim[n = oo][ ( ( 6n^{p}+10 )/( 4n^{p+k}+10 ) ) ] = (1/4)·0^{k+1}

Demostración:

... lim[n = oo][ ( ( 13n^{p}+30 )/( 17n^{p+k}+30 ) ) ] = ...

... lim[n = oo][ ( n^{p}/n^{p} )·( ( 13n^{p}+30 )/( 17n^{p+k}+30 ) ) ]

... lim[n = oo][ ( ( 13+( 30/n^{p} ) )/( 17n^{k}+( 30/n^{p} ) ) ) ] = ( (13+0)/(17·oo^{k}+0) )...

... (13/17)·(1/oo)^{k} = (13/17)·0^{k} = ( (2·6+1)/17 )·0^{k} = (1/17)·0^{k}